Dimension estimates for certain sets of infinite complex

We prove upper and lower estimates on the Hausdorff dimension of sets of infinite complex continued fractions with finitely many prescribed Gaussian integers. Particulary we will conclude that the dimension of theses sets is not zero or two and there are such sets with dimension greater than one and smaller than one.


Introduction
Continued fractions were studied in number theory since the work of Wallis in the 17th century, see [7]. The first dimensional theoretical perspective on infinite real continued fractions can be found in the work of Jarnik [12], who introduced upper and lower estimates on the Hausdorff dimension of sets of continued fractions with bounded digits. The problem of calculating the dimension of these sets has been addressed by several authors [6,1,2,8,9]. In a resent work Jenkinson and Pollicott provide a fast algorithm to approximate this dimension [13]. Dimension theoretical aspects of infinite complex continued fractions were studied by Gardner, Mauldin and Urbanski [5,16]. They proofed that the set of complex continued fractions with arbitrary Gaussian integers from N + Zi has Hausdorff dimension greater than one and smaller than two. We consider here infinite complex continued fractions and ask for the Hausdorff dimension of the set of continued fractions with digits coming from a finite set A ⊂ N[i]. Using the Moran formula from the theory of iterated function systems [15] we are able to give upper and lower estimates on the Hausdorff dimension of these sets, see theorem 2.1. We will show that the dimension of the sets is not zero or two and there are such sets with dimension grater than one and smaller than one, see corollary 2.1 and 2.2. In addition we provide explicit estimates in selected examples.
Given a sequence z n ∈ C for n ∈ N 0 of Gaussian integers we define the infinite complex continued fraction by It is well known that every complex number can be represented as an infinite continued fractions of Gaussian integers using the Hurwitz algorithm [10]. Now fix a finite set We consider the set of all infinite continued fractions having fractional entries coming from A, Obviously the set C(A) is uncountable and it is a null set with respect to the two-dimension Lebesgue measure (this is immediate from corollary 2.1 below). Thus we are interested in the Hausdorff dimension of this set. Recall [3,17] that the d-dimensional Hausdorff measure of a set C ⊆ C is The Hausdorff dimension of C is given by Now we are able to sate our main result on dim H C(A).
With an additional argument this theorem as the following corollary: Hence D < 2. If A has more than one element we have The result now follows from our theorem.
By a similar argument we get the second corollary.
Hence for a suitable choice of A we have For this set A we have d > 1. On the other hand consider A = {1 + i, 2 + i}. We have hence D < 1. The result again follows from our theorem. We remark that it is possible deduce the last corollaries from theorem 1 and theorem 2 of [5] by a few additional arguments. To obtain these results from our main theorem seems to us more transparent. Our last corollary gives the obvious explicit upper and lower bounds following from theorem 2.1: An elementary calculation shows that theorem 2.1 gives 1 < dim H C(A) < 1.33. We like to remark here that it is possible to find an algorithm using thermodynamic formalism that approximate the dimension of C(A). We could apply the recent approach of Jekinsion and Policott [14] to infinite complex continues fractions. This approach has the disadvantage that it is not possible to perform necessary calculations without using a computer, which would change the field of our research to computational mathematics.

Proof of the result
For (a, b) ∈ N 2 consider transformations T a,b : C −→ C given by  Proof. For I(z) = 1/z we have if |z| = r. Applying the translation with a + bi we obtain T a,b (B r (z)) = B r/(|z+a+bi| 2 −r 2 ) ( z + a + bi |z + a + bi| 2 − r 2 ) if |z + a + bi| = r. Especially we get T a,b (B 1/2 (1/2)) = B 1 2a+2a 2 +2b 2 ( 1/2 + a − bi a + a 2 + b 2 ). since |1/2 + a + bi| > 1/2 for a, b > 0. We have to show the distance of the center of the image to 1/2 plus the radius of the image is less or equal to 1/2. This means which is obviously true for a a ∈ N. Next we show that the images of the open balls B 1/2 (1/2) under different T a,b are disjoint.
Proof. We have to show that the distance of the balls at hand is bigger or equal to the sum of there radii, this is With d 1 = a 1 + a 2 1 + b 2 1 and d 2 = a 2 + a 2 2 + b 2 2 we have to show This is obviously true under our assumption.
The last lemma contains estimates on the modulus of derivative of the maps on the closed ballB 1/2 (1/2).
Proof. For z ∈B 1/2 (1/2) we have Now the first estimate is obvious. For the second part note that for all z 1 , z 2 ∈B 1/2 (1/2) and all a + bi ∈ A. Now theorem 2.1 is a direct application of theorem 8.8 of Falconer [4], a well know result in the dimension theory of IFS, which goes back to Moran [15].